📝 题目
例 6 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续,在(a, b)内可导,其中 $a > 0$ . 求证:
(1)存在 $\xi \in \left( {a,b}\right)$ ,使得 $f\left( b\right) - f\left( a\right) = \xi {f}^{\prime }\left( \xi \right) \ln \frac{b}{a}$ ；
(2) $\mathop{\lim }\limits_{{n \rightarrow \infty }}n\left( {\sqrt[n]{x} - 1}\right) = \ln x$ .
💡 答案与解析
证(1)由柯西中值定理, $\exists \xi \in \left( {a,b}\right)$ ,使得
$$ \frac{f\left( b\right) - f\left( a\right) }{\ln b - \ln a} = \frac{{f}^{\prime }\left( \xi \right) }{\frac{1}{\xi }} = \xi {f}^{\prime }\left( \xi \right) $$
$$ \Rightarrow f\left( b\right) - f\left( a\right) = \xi {f}^{\prime }\left( \xi \right) \ln \frac{b}{a}. $$
(2)对 $\forall x > 0$ ,当 $x = 1$ 时,结论显然成立. 当 $x \neq 1$ 时,令
$$ a = \min \{ 1,x\} ,\;b = \max \{ 1,x\} , $$
在 $\left\lbrack {a,b}\right\rbrack$ 上对函数 $f\left( t\right) = {t}^{\frac{1}{n}}$ 利用第 (1) 小题结果,则有 $\exists \xi \in \left( {a,b}\right)$ ,
使得
$$ {b}^{\frac{1}{n}} - {a}^{\frac{1}{n}} = \xi \frac{1}{n}{\xi }^{\frac{1}{n} - 1}\ln \frac{b}{a} $$
$$ \Rightarrow {x}^{\frac{1}{n}} - 1 = \frac{1}{n}{\xi }^{\frac{1}{n}}\ln x \Rightarrow n\left( {{x}^{\frac{1}{n}} - 1}\right) = {\xi }^{\frac{1}{n}}\ln x. $$
因此 $\mathop{\lim }\limits_{{n \rightarrow \infty }}n\left( {\sqrt[n]{x} - 1}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\xi }^{\frac{1}{n}}\ln x = \ln x$ .
\subsubsection{二、用辅助区间法}